Plot triangle ABC with vertices A(2,2), B(4,7), and C(9,2). Find the image of triangle ABC under an anticlockwise rotation of 90 degrees about the origin, and then under a translation by the vector (-

• The scale is 2cm to 2 units on each axis, which means 1cm represents 1 unit.
• The x-axis ranges from $-10$ to 10.
• The y-axis ranges from $-10$ to 10.
• Draw two perpendicular axes (x-axis and y-axis) on graph paper, marking units according to the scale.

The coordinates of triangle ABC are given as $A(2,2)$, $B(4,7)$, and $C(9,2)$.
Plot these points on the graph and connect them to form triangle ABC.

The rule for a $90^\circ$ anticlockwise rotation about the origin $(0,0)$ is $(x,y) \to (-y,x)$.
Applying this rule to the vertices of triangle ABC:

• $A(2,2) \to A'(-2,2)$
• $B(4,7) \to B'(-7,4)$
• $C(9,2) \to C'(-2,9)$ Plot these new points $A'$, $B'$, $C'$ and connect them to form the image triangle $A'B'C'$. The coordinates of the image are:

The rule for a translation by vector $\begin{pmatrix} a \\ b \end{pmatrix}$ is $(x,y) \to (x+a, y+b)$. Here, $a=-10$ and $b=-9$.
Applying this rule to the vertices of triangle ABC:

• $A(2,2) \to A''(2+(-10), 2+(-9)) = A''(2-10, 2-9) = A''(-8,-7)$
• $B(4,7) \to B''(4+(-10), 7+(-9)) = B''(4-10, 7-9) = B''(-6,-2)$
• $C(9,2) \to C''(9+(-10), 2+(-9)) = C''(9-10, 2-9) = C''(-1,-7)$ Plot these new points $A''$, $B''$, $C''$ and connect them to form the image triangle $A''B''C''$. The coordinates of the image are:

The rule for a reflection in the x-axis ($y=0$) is $(x,y) \to (x,-y)$.
Applying this rule to the vertices of triangle ABC:

• $A(2,2) \to A'''(2,-2)$
• $B(4,7) \to B'''(4,-7)$
• $C(9,2) \to C'''(9,-2)$ Plot these new points $A'''$, $B'''$, $C'''$ and connect them to form the image triangle $A'''B'''C'''$. The coordinates of the image are: